Number of Men in the Desert

Here is a problem by J.A.H. Hunter.

'I don't get this,' Tony looked up from his book. 'It says each man carried enough to feed one man seven days, and that was as much as any of them could carry. They picked up no food out in the desert, but the leader was away from their base 12 days.' His father laughed, 'I read that, too,' he told the boy, 'and it's explained later on. They all started back separately, and only the leader went as far as six days each way.' Tony pondered this. 'I see,' he said at last. 'They figured it so that the last man could be out as long as possible, but without setting up a store of grub any place.' That's just what they had done. So how many men started?

Please give the answer and the method to get the answer.

Peter Heichelheim

SPOILER . . . . . . . . . . . . . Six(6) men started the trek.

At the end of day one(1), the first man had consumed one day of supplies. He then turned over one day of supplies to each of the five remaining men and used his remaining day of food to return to base. Similarly, for the second man at the end of day two(2), the third man at the end of day three(3) .... etc.

m1 1 day + 1 day back + 5*1 day

Alternatively you could set out with each person only feeding the leader, and passing on parcel/s of food they did not need to return home, to someone else left behind, ie the first person to drop off may feed the leader on the first day, depart on the second day with one, while passing on the extra (4) parcels to everybody else. (7-1-1-1)=4 which is how many people have space for one more bundle.

Very good. You got J.A.H. Hunter's answer which was 6 men started. Anybody able to show less men could do it without anyone making a multiple trip or prove it it is impossible.

How about some different assumptions: 1) Can I leave supplies in the desert, and count finding them when I get back? 2) Can I travel for a day without supplies, if I have them at the end of the day's march. For example, can I walk for a day without supplies, if I can count on finding them where I left them, or if I am out of the desert by the end of the day?

No, that would be a different problem. It says 'without setting up a store of grub any place' and there is an implied presumption they would actually use the food they were carrying.

how many people are needed to keep the leader walk in the desert for 22 days?

answer: 66 incl the leader, each person goes out only once.

rove it it is impossible. : eter

6 men (incl the leader) are the least to make this, but we can let the leader start alone.

it is absolutely alright if you say this is a trick. ;^)

6 - L(1) 5 - L(2) LM(2) 4 - L(3) M(3) LMN(3) dist.3 - L(4) M(4) N(4) LMNO(4) from 2 - L(5) M(5) N(5) O(5) LMNOP(5) 1 - L(6) M(6) N(6) O(6) P(6) LMNOPQ(6) CAMP 0

Why stop at only 6 days out? The leader can almost go out 7 days each way....

-danny

:> > Here is a problem by J.A.H. Hunter. :> > :> > 'I don't get this,' Tony looked up from his book. 'It says each man :> > carried enough to feed one man seven days, and that was as much as any :> > of them could carry. They picked up no food out in the desert, but the :> > leader was away from their base 12 days.' :> > His father laughed, 'I read that, too,' he told the boy, 'and it's :> > explained later on. They all started back separately, and only the :> > leader went as far as six days each way.' :> > Tony pondered this. 'I see,' he said at last. 'They figured it so :> > that the last man could be out as long as possible, but without :> > setting up a store of grub any place.' :> > That's just what they had done. So how many men started? :> > :> SPOILER :> . :> . :> . :> . :> . :> . :> . :> . :> . :> . :> . :> . :> . :> Six(6) men started the trek. :> :> At the end of day one(1), the first man had consumed one day of supplies. :> He then turned over one day of supplies to each of the five remaining men :> and used his remaining day of food to return to base. Similarly, for the :> second man at the end of day two(2), the third man at the end of day :> three(3) .... etc. :> :> m1 1 day + 1 day back + 5*1 day

I should add that d(k) is the distance travelled INTO the desert by the kth member. The total distance travelled by him would 2d(k).

Upon further investigation, I noticed that d(1)=d(2)=...=d(N). Using this result, the farthest INTO the desert one member of the team can go is D(N)=NM/(N+1) From this, N=D(N)/[M-D(N)] When D(N)=6 and M=7, it's easy to see that N=6.